Mathematical Rules of Thumb, illustrated reader · Chapter 18

18Applied Mathematics

Scaling, Regimes, and Model Sanity

4 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 18 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-applied-mathematics from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 4

Screen a lumped thermal model

Does a computed time constant validate the material properties?

Compare the internal-conduction scale with surface heat transfer before accepting a uniform-temperature model.

Bi=hLck,Lc=r/3,τ=ρcr3h Bi=\frac{hL_c}{k},\quad L_c=r/3,\quad\tau=\frac{\rho c r}{3h}

Heat transfer h (W/m²/K). Sphere radius .01m; k=200W/m/K, ρ=2700kg/m³, c=900J/kg/K. Small Bi is a screening condition.

Predict first. Does a computed time constant validate the material properties?

Choose an example

Screen a lumped thermal model. Lc=r/3 gives Bi=0.00333333. The screen passes, so the lumped time constant τ=ρcLc/h=40.5 s is a reasonable estimate under assumed aluminium-like properties. The .1 screen is a rule of thumb; it does not validate real material properties or external heat transfer.
Heat transfer h (W/m²/K): 200
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Radius (m)
0.01
Conductivity (W/m/K)
200
Density (kg/m³)
2700
Specific heat (J/kg/K)
900
Heat transfer h (W/m²/K)
200
Biot number
0.00333333
Screen Bi < .1 passed
True
Lumped time constant (s)
40.5

Lc=r/3 gives Bi=0.00333333. The screen passes, so the lumped time constant τ=ρcLc/h=40.5 s is a reasonable estimate under assumed aluminium-like properties. The .1 screen is a rule of thumb; it does not validate real material properties or external heat transfer.

Use the idea

Use rule 18.3.5 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Sphere radius .01m; k=200W/m/K, ρ=2700kg/m³, c=900J/kg/K. Small Bi is a screening condition.

Check your understanding: Does a computed time constant validate the material properties?
No. It is conditional on the assumed properties, geometry and applicable lumped model.

Book source: Rule 18.3.5: Use lumped thermal capacitance only for small Biot number. Demonstration C18-D01. Worked illustration.

2Demonstration 2 of 4

Compare transport time scales

What does Pe≫1 indicate here?

The diffusion and advection time curves cross at a characteristic length.

tD=L2/D,tA=L/U,Pe=UL/D t_D=L^2/D,\quad t_A=L/U,\quad Pe=UL/D

Transport speed U (m/s). D=.01m²/s, positive speed and length; regime comparison does not supply boundary conditions.

Predict first. What does Pe≫1 indicate here?

Choose an example

Compare transport time scales. At L=.1 m, Pe=UL/D=1. Flow and diffusion take equal time here. Boundaries and geometry still shape the details.
Transport speed U (m/s): 0.1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Speed U (m/s)
0.1
Diffusivity D (m²/s)
0.01
Peclet at L=.1m
1
Equal-time length (m)
0.1

At L=.1 m, Pe=UL/D=1. Flow and diffusion take equal time here. Boundaries and geometry still shape the details.

Use the idea

Use rule 18.3.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

D=.01m²/s, positive speed and length; regime comparison does not supply boundary conditions.

Check your understanding: What does Pe≫1 indicate here?
Advection acts over the length scale faster than diffusion; it does not by itself determine every feature of the solution.

Book source: Rule 18.3.2: Use Peclet number to compare advection with diffusion. Demonstration C18-D02. Worked illustration.

3Demonstration 3 of 4

Construct a dimensionless flow comparison

Can one Reynolds cutoff classify all geometries?

Use matched units to compare inertia with viscous effects at a fixed physical scale.

Re=ρULμ Re=\frac{\rho U L}{\mu}

Speed U (m/s). ρ=1000kg/m³, L=.01m, μ=.001Pa s. Flow transition depends on geometry and disturbances.

Predict first. Can one Reynolds cutoff classify all geometries?

Choose an example

Construct a dimensionless flow comparison. With the disclosed density, viscosity and length, U=0.1 m/s gives Re=1000. Inertia outweighs viscosity, yet many flows are still smooth at this size; pipe flow, for example, usually stays laminar below about 2000. The exact switch point depends on geometry, so no single cutoff fits every flow.
Speed U (m/s): 0.1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Density (kg/m³)
1000
Length (m)
0.01
Dynamic viscosity (Pa s)
0.001
Selected Reynolds number
1000

With the disclosed density, viscosity and length, U=0.1 m/s gives Re=1000. Inertia outweighs viscosity, yet many flows are still smooth at this size; pipe flow, for example, usually stays laminar below about 2000. The exact switch point depends on geometry, so no single cutoff fits every flow.

Use the idea

Use rule 18.3.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

ρ=1000kg/m³, L=.01m, μ=.001Pa s. Flow transition depends on geometry and disturbances.

Check your understanding: Can one Reynolds cutoff classify all geometries?
No. A threshold adopted for one flow arrangement is not a universal turbulence rule.

Book source: Rule 18.3.1: Use Reynolds number to compare inertia with viscosity. Demonstration C18-D03. Worked illustration.

4Demonstration 4 of 4

Feel the square law of diffusion time

How much longer does a 4 cm slab take than a 2 cm slab?

Diffusion time grows with the square of distance, so thickness matters more than intuition says.

Fo=αtL2≈1,t≈L2α Fo=\frac{\alpha t}{L^2}\approx1,\quad t\approx\frac{L^2}{\alpha}

Slab thickness L (m). Thermal diffusivity α=1e-5 m²/s (steel-like); order-of-magnitude estimate at Fourier number 1.

Predict first. How much longer does a 4 cm slab take than a 2 cm slab?

Choose an example

Feel the square law of diffusion time. Heat diffuses through 2 cm in about L²/α=40 s (Fourier number 1). This is the reference slab. Time grows with the square of distance because diffusion is a random walk.
Slab thickness L (m): 0.02
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Thickness (cm)
2
Thermal diffusivity (m²/s)
1e-05
Penetration time (s)
40
Time vs 2 cm slab
1

Heat diffuses through 2 cm in about L²/α=40 s (Fourier number 1). This is the reference slab. Time grows with the square of distance because diffusion is a random walk.

Use the idea

Use rule 18.3.4 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Thermal diffusivity α=1e-5 m²/s (steel-like); order-of-magnitude estimate at Fourier number 1.

Check your understanding: How much longer does a 4 cm slab take than a 2 cm slab?
Four times as long: 160 s versus 40 s.

Book source: Rule 18.3.4: Use Fourier number to estimate diffusion penetration time. Demonstration C18-D04. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.